They are the element symbols. Too many to list here, look up a periodic table.
Answer:
Explanation:
Suppose mass of block 1 is
and
of block 2
For original system
natural frequency of oscillation is given by
![\omega _1=\sqrt{\frac{k}{m_1}}](https://tex.z-dn.net/?f=%5Comega%20_1%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm_1%7D%7D)
Maximum kinetic Energy is equal to total Energy of the system
![K=\frac{1}{2}kA^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7DkA%5E2)
where k=spring constant
A=maximum amplitude
Now Block B is Placed at block 1 at maximum amplitude such that A=A'
i.e. new amplitude=old amplitude
Maximum kinetic energy of combined system is
![K=\frac{1}{2}kA'^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7DkA%27%5E2)
as the total energy is independent of mass therefore maximum kinetic energy will remain same
Answer:
The planes´s horizontal distance is DH = 13845 meters
Explanation:
We need to get the vector components because we already know the distance and the angle so using the trigonometric identity of Cos(Ф) we can get the horizontal component and if we get the component on x axle like 22000 (m)*Cos(51°) = x so we can find that x = 13845.05 meters
The work done by a force is equal to:
![W=Fd](https://tex.z-dn.net/?f=W%3DFd)
where
W is the work
F is the force applied
d is the distance through which the force is applied.
In our problem, the force needed to lift the dog is equal to the weight of the dog:
F=24 N
and the dog is moved through a distance of
d=4.8 m
therefore, the work done by the child to lift the dog is
Answer:
A,
Explanation:
To solve the exercise it is necessary to use the concepts and definitions given for electromagnetic energy, in which
![E = \frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
Where,
h = Plank's constant ![(6.626*10^{-34}J.s)](https://tex.z-dn.net/?f=%286.626%2A10%5E%7B-34%7DJ.s%29)
c = Speed of light ![(3*10^8m/s)](https://tex.z-dn.net/?f=%283%2A10%5E8m%2Fs%29)
=Wavelength
If we analyze these characteristics both h and c are constant so the energy is inversely proportional to the size of the wave.
The larger the amplitude of the wave, the smaller the energy.
On the other hand we have the frequency value defined as
![f = \frac{c}{\lambda}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D)
In this case the frequency is also inversely proportional to the wavelength.
In this case, the amplitude of the largest wave is infrared, so it will have less energy and less frequency. The fact that it has a low frequency by the wavelength, also generates that it has a low energy. But not because it has a large wavelength, on the contrary, because its wavelength is small.
In the case of the ultraviolet wave it will have greater frequency and greater energy. Therefore of all the options, A is the only one valid.